log-likelihood for the Poisson distribution
Arguments
- x
non negative integers
- lambda
non-negative means
- full
Add the data frame showing x, mean, sd as well as the fx and derivatives
Value
data frame with fx for the pdf value of with
dLambda that has the derivatives with respect to the parameters at
the observation time-point
Details
In an rxode2() model, you can use llikPois() but you have to
use all arguments. You can also get the derivatives with
llikPoisDlambda()
Examples
# \donttest{
llikPois(0:7, lambda = 1)
#> fx dLambda
#> 1 -1.000000 -1
#> 2 -1.000000 0
#> 3 -1.693147 1
#> 4 -2.791759 2
#> 5 -4.178054 3
#> 6 -5.787492 4
#> 7 -7.579251 5
#> 8 -9.525161 6
llikPois(0:7, lambda = 4, full=TRUE)
#> x lambda fx dLambda
#> 1 0 4 -4.000000 -1.00
#> 2 1 4 -2.613706 -0.75
#> 3 2 4 -1.920558 -0.50
#> 4 3 4 -1.632876 -0.25
#> 5 4 4 -1.632876 0.00
#> 6 5 4 -1.856020 0.25
#> 7 6 4 -2.261485 0.50
#> 8 7 4 -2.821101 0.75
# In rxode2 you can use:
et <- et(0:10)
et$lambda <- 0.5
model <- function() {
model({
fx <- llikPois(time, lambda)
dLambda <- llikPoisDlambda(time, lambda)
})
}
rxSolve(model, et)
#>
#>
#>
#>
#> using C compiler: ‘gcc (Ubuntu 11.4.0-1ubuntu1~22.04) 11.4.0’
#> ── Solved rxode2 object ──
#> ── Parameters ($params): ──
#> # A tibble: 1 × 0
#> ── Initial Conditions ($inits): ──
#> named numeric(0)
#> ── First part of data (object): ──
#> # A tibble: 11 × 4
#> time fx dLambda lambda
#> <dbl> <dbl> <dbl> <dbl>
#> 1 0 -0.5 -1 0.5
#> 2 1 -1.19 1 0.5
#> 3 2 -2.58 3 0.5
#> 4 3 -4.37 5 0.5
#> 5 4 -6.45 7 0.5
#> 6 5 -8.75 9 0.5
#> # ℹ 5 more rows
# }